The amount earned on the first day is Rs.200.
Let the amount spent on the first day be Rs. x. Amount with her at the end of the first day is (200 − x).
Amount earned on the second day is Rs.300 and the amount spent on the second day is Rs.2x. The amount left on the second day is 300 − 2x.
Similarly, the net amount that she would have on the third day is 400 − 4x. Therefore, the total amount that she would have at the end of three days is (200 − x) + (300 − 2x) + (400 − 4x).
That is, 200 + 300 + 400 + ( − 1 − 2 − 4)x
= 900 + ( −7)x
Thus the required algebraic expression is 900 − 7x.

i) (3x − 4y + z) + (2x − z + 3y)
= (3x + 2x) + ( −4y + 3y) + (z − z)
= (3 + 2)x + ( −4 + 3)y + (1 − 1)z
= 5x − 1y + 0z
= 5x − y.

ii) (4x + 3y) − (2x − 5y)
= (4x + 3y) + ( −2x + 5y)
= (4x − 2x) + (3y + 5y)
= (4 − 2)x + (3 + 5)y
= 2x + 8y.

In the given algebraic expression, x,y,z are the variables.
Let us group the like terms.
100x + 99y − 98z + 10x + 10y + 10z − x − y + z
= (100x + 10x − x) + (99y + 10y − y) + ( −98z + 10z + z)
= (100 + 10 − 1)x + (99 + 10 − 1)y + ( −98 + 10 + 1)z
= (110 − 1)x + (109 − 1)y + ( −98 + 11)z
= 109x + 108y + ( −87)z
= 109x + 108y − 87z.

i) 11pq – 7pq. Additive inverse of 7pq is − 7pq
11pq + ( − 7pq) = 11pq − 7pq = (11 − 7)pq = 4pq
ii) a – (–a). Additive inverse of −a is a.
So, a + a = 2a
iii) 21x + 9 – (5x + 7). Additive inverse of 5x + 7 is – (5x + 7).
(21x + 9) + [– (5x + 7)] = (21x + 9) – (5x + 7)
= 21x + 9 − 5x − 7
= (21 − 5)x + (9 − 7)
= 16x + 2.

i) (pq − 1) + (3pq + 2) = (pq + 3pq) + ( − 1 + 2)
= (1 + 3)pq + 1
= 4pq + 1
ii) (8x + 3) + (1 − 7x) = 8x + 3 + 1 − 7x
= (8x − 7x) + (3 + 1)
= (8 − 7) x + 4
= x + 4.